Question

我正在编写SML代码，根据具体公式计算一段时间内的活人，死人和僵尸的数量。我已将我的代码放入“.sml”文件中，然后在SMLNJ中打开它。我得到的错误如下。

zombies.sml:36.3 Error: syntax error: inserting LPAREN
zombies.sml:43.1 Error: syntax error found at EOF

这是我的代码。

val living = 500000.0;
val zombie = 0.0;
val dead = 1.0;

val counter = 0.0;

val d_ = 0.0001; (*chance of natural death*)
val z_ = 0.0001; (*chance of zombification after death*)
val b_ = 0.0095; (*chance of zombie attack*)
val a_ = 0.005; (*chance of a zombie dying?*)

val birth_rate = 1.0;

(*calculates amount of living*)
fun S(living:real, birth_rate:real, b_:real, zombie:real, d_:real) = birth_rate - b_*living*zombie - d_*living;

(*calculates amount of zombies*)
fun Z(zombie:real, b_:real, living:real, z_:real, dead:real, a_:real) = b_*living*zombie + z_*dead - a_*living*zombie;

(*calculates amount of dead*)
fun R(dead:real, d_:real, living:real, a_:real, zombie:real) = d_*living + a_*living*zombie - zombie*dead;

fun program (living:real, zombie:real, dead:real, birth_rate:real, b_:real, z_:real, d_:real, a_:real, counter:real) =
= if counter = 0 then 1
= else (S(living, birth_rate, b_, zombie, d_);
        Z(zombie, b_, living, z_, dead, a_);
        R(dead, d_, living, a_, zombie);
        counter = (counter + 1);
        program(living, zombie, dead, birth_rate, b_, z_, d_, a_, counter));

program(living, zombie, dead, birth_rate, b_, z_, d_, a_, counter);

最后，工作编辑。功能语言很奇怪。

fun add(x, y) = x + y;

(*calculates amount of living*)
fun S(living, birth_rate, b_, zombie, d_) =
round(real(birth_rate+living) - b_*real(living)*real(zombie) - d_*real(living));

(*calculates amount of zombies*)
fun Z(zombie, b_, living, z_, dead, a_) = 
round(b_*real(living)*real(zombie) + z_*real(dead) - a_*real(living)*real(zombie));

(*calculates amount of dead*)
fun R(dead, d_, living, a_, zombie) =
round(d_*real(living) + a_*real(living)*real(zombie) - real(zombie*dead));

fun main (living) =
let
    val living = ref living;
    val zombie = ref 1;
    val dead = ref 1;
    val birth_rate = 1;
    val d_ = 0.0001; (*chance of natural death*)
    val z_ = 0.0001; (*chance of zombification after death*)
    val b_ = 0.0095; (*chance of zombie attack*)
    val a_ = 0.005; (*chance of a zombie dying?*)
    val count = ref 1;
in
    while (!count < 10) do
    (
        living := S(!living, birth_rate, b_, !zombie, d_);
        print("Living count: " ^ Int.toString(!living) ^ "\n");
        zombie := Z(!zombie, b_, !living, z_, !dead, a_);
        print("Zombie count: " ^ Int.toString(!zombie) ^ "\n");
        dead := R(!dead, d_, !living, a_, !zombie);
        print("Dead count: " ^ Int.toString(!dead) ^ "\n");
        count := !count + 1;
        print(Int.toString(!count) ^ "\n")
    )
end;

Answer 1

该行开头的那些=显然是假的：

fun program (…) =
= if counter = 0 then 1
= else (S(living, birth_rate, b_, zombie, d_);

也许文件在传输中被破坏了？（=用作MIME中的行继续符，但它位于行的末尾。）

此外，整数和浮点数是ML中的不同类型。（这也适用于文字。）您需要决定counter和birth_rate整数或实数。

该行

counter = (counter + 1);

没有任何影响。变量在SML中是不可变的。您需要使用let绑定，或者直接使用递增的计数器值调用program。

Answer 2

要在StackOverflow上询问可能为useful for others的问题，请参阅此部分：

有些问题仍然偏离主题，即使它们符合上述类别之一：


问题寻求调试帮助（“为什么这段代码不起作用？”）必须包含所需的行为，特定的问题或错误以及在问题本身中重现它所需的最短代码。没有明确问题陈述的问题对其他读者没有用。请参阅：How to create a Minimal, Complete, and Verifiable example。

在您的示例中，您已经确定问题出在代码的最后部分，但由于最后一部分取决于第一部分，您不能简单地发布最后一部分，因为否则问题可以不被复制。但语法错误可以。

在你达到这一点之前减少直线：

fun program (living:real, zombie:real, dead:real, birth_rate:real, b_:real, z_:real, d_:real, a_:real, counter:real) =
= if counter = 0 then 1
= else raise Fail "Not relevant"

会产生同样的错误。

也许你最好使用一个REPL来提供比SML / NJ更好的错误信息。

比较例如SML / NJ的输出：

typo.sml:24.3 Error: syntax error: inserting  LPAREN
typo.sml:33.1 Error: syntax error found at EOF
/usr/lib/smlnj/bin/sml: Fatal error -- Uncaught exception Compile with "syntax error" raised at ../compiler/Parse/main/smlfile.sml:15.24-15.46

与Moscow ML：

File "typo.sml", line 24, characters 2-4:
! = if counter = 0 then 1
!   ^^
! Syntax error.

在SML中插入LPAREN错误

2 个答案: